3. * The force of interest 8(t) at time t (measured in years) is a +bt² where a and b are constants. An amount of £200 at time t = 0 accumulates to £210 at t = 5 and £230 at t = 10. (a) Show that a = b - log (1.05)-log(1.15) 1 20 log (1.15) - 125 log (1.05) = 0.0001687. = 0.008352, = and (b) Compute A(0,7) and hence compute the discounted value at t = £750 due at t = 7. 0 of a payment of (c) Compute A (6,7). What is the equivalent constant annual interest rate for the year from t 6 to t = = 7? (d) Calculate the constant force of interest that would give rise to the same accumulation from t = 0 to t = 10.
3. * The force of interest 8(t) at time t (measured in years) is a +bt² where a and b are constants. An amount of £200 at time t = 0 accumulates to £210 at t = 5 and £230 at t = 10. (a) Show that a = b - log (1.05)-log(1.15) 1 20 log (1.15) - 125 log (1.05) = 0.0001687. = 0.008352, = and (b) Compute A(0,7) and hence compute the discounted value at t = £750 due at t = 7. 0 of a payment of (c) Compute A (6,7). What is the equivalent constant annual interest rate for the year from t 6 to t = = 7? (d) Calculate the constant force of interest that would give rise to the same accumulation from t = 0 to t = 10.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:3. * The force of interest 8(t) at time t (measured in years) is a +bt² where a and b are constants.
An amount of £200 at time t = 0 accumulates to £210 at t = 5 and £230 at t = 10.
(a) Show that
a =
b
t
log (1.05) -log(1.15) = 0.008352,
30
log(1.15) -log(1.05)
0.0001687.
=
1
=
250
=
(b) Compute A(0,7) and hence compute the discounted value at t =
£750 due at t = 7.
(c) Compute A(6,7). What is the equivalent constant annual interest rate for the year
from t = 6 to t = 7?
and
(d) Calculate the constant force of interest that would give rise to the same accumulation
from 0 to t = 10.
0 of a payment of
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